Question

Does `lim_(xrarroo) n^x = oo`?

Answer 1

Infinity is not a number. But see below.

Explanation:

For `n > 1`
`color(white)("xxx")lim_(xrarroo)n^x = oo`

For `0 < n < 1`
`color(white)("xxx")lim_(xrarroo)n^x = 0`

For `n=1`
`color(white)("xxx")lim_(xrarroo)1^x = lim_(xrarroo) 1 = 1`
This is true only if the base really is `1` and not some function that approaches `1` as `x` increases without bound (as `xrarroo)`

Is the base is a function that approaches `1` as `xrarroo`, then the form `1^oo` is indeterminate.

That is:

If `lim_(xrarroo)g(x) = 1` and `lim_(xrarroo)f(x) = oo`, then

`lim_(xrarroo)(g(x))^f(x)` is indeterminate.

For example `lim_(xrarroo)(1+1/x)^x = e` and `lim_(xrarroo)(1+5/x)^x = e^5`